Solution :
Given :
Initial temperature of the refrigerant is :
= ( 39.37 + 273 ) K
= 312.3 K
Room which is maintained at constant temperature is :
= (22+273) K
= 295 K
The thermal energy transferred to the room is :
Q = 400 kJ
= 
Therefore, the total entropy generation during the thermal energy process is :
![$\Delta S =\left[\frac{-Q}{T_i}+ \frac{+Q}{T_i}\right]$](https://tex.z-dn.net/?f=%24%5CDelta%20S%20%3D%5Cleft%5B%5Cfrac%7B-Q%7D%7BT_i%7D%2B%20%5Cfrac%7B%2BQ%7D%7BT_i%7D%5Cright%5D%24)
Here, -Q = heat is leaving the system maintained at a temperature of 
K.
+Q = heat is entering the system maintained at a temperature of 
K.
Therefore, substituting the values :
![$\Delta S =\left[\frac{-400\times 10^3}{312.3}+ \frac{400\times 10^3}{295}\right]$](https://tex.z-dn.net/?f=%24%5CDelta%20S%20%3D%5Cleft%5B%5Cfrac%7B-400%5Ctimes%2010%5E3%7D%7B312.3%7D%2B%20%5Cfrac%7B400%5Ctimes%2010%5E3%7D%7B295%7D%5Cright%5D%24)
= [-1280.8197 + 1355.9322]
= 75.1125 J/K
= 0.0751125 kJ/K
= 0.075 kJ/K
Honey is a - Mixture
Ink is - Mixture
And Gold Dust is a - Compound
They’re put in groups that have similar properties to them. but the periodic table is organized by atomic numbers.
The rate of particle movement decreases. A direct effect of this is the decrease in particle distance.
Here we have to calculate the number of moles of valuable propane can be prepared from 1.8 moles of carbon.
From 1.8 moles of carbon 0.3 moles of propane can be prepared by the reaction.
From 6 moles of carbon (C) 1 moles of valuable propane (C₃H₈) can be prepared.
Thus from 1.8 moles of C we can obtain 
×1.8 = 0.3 moles of the propane can be prepared.
Thus the amount of propane produced in this reaction is determined.
